Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $n = \dfrac{-7y - 42}{y^2 - 13y + 42} \times \dfrac{-6y + 42}{y + 6} $
Solution: First factor the quadratic. $n = \dfrac{-7y - 42}{(y - 7)(y - 6)} \times \dfrac{-6y + 42}{y + 6} $ Then factor out any other terms. $n = \dfrac{-7(y + 6)}{(y - 7)(y - 6)} \times \dfrac{-6(y - 7)}{y + 6} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ -7(y + 6) \times -6(y - 7) } { (y - 7)(y - 6) \times (y + 6) } $ $n = \dfrac{ 42(y + 6)(y - 7)}{ (y - 7)(y - 6)(y + 6)} $ Notice that $(y + 6)$ and $(y - 7)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ 42(y + 6)\cancel{(y - 7)}}{ \cancel{(y - 7)}(y - 6)(y + 6)} $ We are dividing by $y - 7$ , so $y - 7 \neq 0$ Therefore, $y \neq 7$ $n = \dfrac{ 42\cancel{(y + 6)}\cancel{(y - 7)}}{ \cancel{(y - 7)}(y - 6)\cancel{(y + 6)}} $ We are dividing by $y + 6$ , so $y + 6 \neq 0$ Therefore, $y \neq -6$ $n = \dfrac{42}{y - 6} ; \space y \neq 7 ; \space y \neq -6 $